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sergiy2304 [10]
3 years ago
15

How is the basic structure of diamond related to its bulk properties?

Physics
1 answer:
KATRIN_1 [288]3 years ago
4 0
Diamonds structure is defined by series of multiple covalent bonds. Given that it is made of a very strong bond, this makes some of its bulk properties relatable to this. Example its density which is larger than most of the crystalline. Another is its intensive property of its temperature, given that the diamond is hard to break down, it also has high melting point.
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Which of the following gases are the heaviest? <br> O2, CH4, CO2, Cl2
kvasek [131]

the answer is

CI2 because its 70.


5 0
3 years ago
A gas station owner suspects that he is being overcharged for gasoline deliveries by a gasoline supplier. The overcharge seems p
Tems11 [23]

Answer:

Explanation:

delta V = v * alpha * delta T

= V * 0.00053 * (92.2 - 55.0)

= 0.019716 V

percentage that the owner

= [delta V / V] * 100

= [0.019716 V / V] * 100

= 1.9716 %

4 0
2 years ago
A 200 g hockey puck is launched up a metal ramp that is inclined at a 30° angle. The coefficients of static and kinetic friction
nikitadnepr [17]

Answer:

71.76 m

Explanation:

We will solve this question using the work energy theorem.

The theorem explains that, the change in kinetic energy of a particle between two points is equal to the workdone in moving the particle from the one point to the other.

ΔK.E = W

In the attached free body diagram for the question, the forces acting on the puck are given.

ΔK.E = (final kinetic energy) - (initial kinetic energy)

Final kinetic energy = 0 J (since the puck comes to a stop)

Initial kinetic energy = (1/2)(m)(v²) = (1/2)(0.2)(26²) = 67.6 J

ΔK.E = 0 - 67.6 = - 67.6 J

W = Workdone between the starting and stopping points = (work done by the force of gravity) + (work done by frictional force)

Work done by the force of gravity = - mgh = - (0.2)(9.8)(h) = - 1.96 h

Workdone by the frictional force = F × d

F = μ N

μ = coefficient of kinetic friction = 0.30 (kinetic frictional force is the only frictional force that moves a distance of d, the static frictional force doesn't move any distance, so it does no work)

N = normal reaction of the plane surface on the puck = mg cos 30° = (0.2)(9.8)(0.866) = 1.697 N

F = μ N = 0.3 × 1.697 = 0.509 N

where d = distance along the incline that the puck travels.

d = h/sin 30° = 2h (from trigonometric relations)

Workdone by the frictional force = F × d = 0.509 × 2h = 1.02 h

ΔK.E = W = (work done by the force of gravity) + (work done by frictional force)

- 67.6 = - 1.96h + 1.02h

-0.942h = - 67.6

h = 71.76 m

6 0
2 years ago
Shondra takes notes in class.
sdas [7]

Answer:

I beileve its A because energy is what gives the ability to do work

Explanation:

4 0
2 years ago
A positively charged particle is in the center of a parallel plate capacitor that has charge +/- Q on it's plates. Suppose the d
ivolga24 [154]

Answer:

the force remains constant if the charge does not change

Explanation:

In a capacitor the capacitance is given by

           C = ε₀ A / d

Where ε₀ is the permissiveness of emptiness, A is about the plates and d the distance between them.

The charge on the capacitor is given by the ratio

            Q = C ΔV

Let's apply these expressions to our problem, if the load remains constant

            C = Q / ΔV = ε₀ A / d

            ΔV / d = Q / ε₀ A

If the distance increases the capacitance should decrease, therefore if the charge is a constant the voltaje difference must increase

Now we can analyze the force on the test charge in the center of the capacitor

               ΔV = E d

               E= ΔV/d

               F = q E

              F = q ΔV / d

 Let's replace

          F = q Q /ε₀ A

From this expression we see that the force is constant since the voltage increase is compensated by increasing the distance, therefore the correct answer is that the force remains constant if the charge does not change

8 0
2 years ago
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